I'm assuming q(x) = √(x + 4)
p(q(x)) = (√(x + 4))² + 7 = x + 4 + 7 = x + 11
p(q(5)) = 5 + 11 = 16
q(p(x)) = √(x² + 7 + 4) = √(x² + 11)
q(p(5)) = √(5² + 11) = √(25 + 11) = √36 = 6
Germ R.
asked • 05/23/17Functions and Function composition problem. Please help with problem
p(x)=x^2+7
q(x)=Sq.Rt. x+4
Find:
(p FunctionComp q)(5) or p(q(5))=
(p FunctionComp q)(5) or q(p(5))=
Thank You!
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2 Answers By Expert Tutors
Andrew M. answered • 05/23/17
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
p(x) = x2+7
q(x) = √(x+ 4)
q(5) = √(5+4) = √9 = {3, -3}
p(q(5)) = p(3) or p(-3)
32 + 7 = 16 or
(-3)2 + 7 = 16
p(q(5))=16
p(5)=52+7 = 32
q(p(5)) = q(32) =
√(32+4) = √36 = {-6, 6}
Michael A.
tutor
Andrew, would you consider the negative square root? √36 implies taking the principal, or positive square root, which is 6.
If you want to consider the negative square root, wouldn't it have to be written as -√36, which says to multiply -1 times the square root of 36, giving us -1 * 6 = -6?
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05/23/17
Andrew M.
Michael,
If I square (-6) do I not get 36?
If I take -√36 then I get -1 times
either 6 or -6
-(6) = -6
-(-6) = -6
The answer is the same. You are
discussing implications while I have
simply stated the mathematical fact.
62 = 6(6) = 36
(-6)2 = (-6)(-6) = 36
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05/24/17
Andrew M.
And, yes, I do see the extra negative sign I
typoed in my comment.
-(6) = -6
-(-6) = 6
Germ,
when solving equations through processes such
as completing the square (which is the method
utilized the derive the quadratic equation from
ax2 + bx +c = 0) the ± is put there to remind
people that numbers such as √36 have TWO
solutions... 6 and -6
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05/24/17
Michael A.
tutor
Andrew, you are absolutely correct. (-6)(-6) = 36 and (6)(6) = 36. That is an undeniable fact. I'm not arguing the point. I am just asking whether the negative should be considered. The range of √x is (0, ∞), for example. In all of my experience, when asked to find the square root of a number such as √36, that meant to find the positive square root, which is 6. If that is no longer the way this is taught, then I'm glad to be made aware of this.
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05/24/17
Andrew M.
Michael;
I agree wholeheartedly that the range of
√x is [0, ∞). That has no true bearing on
the square root of a known positive integer.
What I was looking at was the √36 which
is either 6 or -6. So, yes, I do believe the
negative should be considered. Whether or
not Germ's instructor thinks so we may
never know. Thank you for your input.
Have a great day. :-)
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05/25/17
Michael A.
tutor
Thanks. Same to you. I respect your considerable wealth of knowledge.
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05/25/17
Germ R.
Thank you both for the discussion. I also had questions if the negative should be considered and had contacted the company about it. Some of their problems they ask and answers they require are vague. I call them (Aleks) often to have them clarify what they're looking for. It has happened with other problems as well. Their system is electronic so if the answer is not exactly what they're looking for it is kicked out as wrong. I hope this answers some of your questions. Thank you again.
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05/25/17
Michael A.
tutor
You're welcome, Germ. Thank you, as well. I have prior experience with ALEKS during my time as a pre-algebra and algebra instructor for a community college. You are exactly correct. It will label the answer incorrect if you do not supply exactly what they are seeking. Other times, it will tell you that the answer is acceptable, but it can be written in a better, or different format. ALEKS is a good product, but it can be "persnickety."
Report
05/25/17
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Germ R.
05/23/17